Wave Optics — Questions & Answers

Sample questions from Wave Optics

1. Q1. For CONSTRUCTIVE interference (a bright fringe) to occur at a point, the PATH DIFFERENCE between the two interfering waves arriving at that point must be:

   • A.An integer multiple of half the wavelength specifically
   • B.An integer multiple of the wavelength (0, lambda, 2 lambda, 3 lambda, and so on)✓
   • C.Always exactly zero, with no other possibility
   • D.A completely random, unrestricted value
   Solution

   Constructive interference (producing a bright fringe/maximum) occurs when the PATH DIFFERENCE between the two interfering waves is an INTEGER multiple of the wavelength: path difference = n x lambda, where n=0, 1, 2, 3, and so on -- at such points, the two waves arrive exactly IN PHASE, their amplitudes adding constructively to produce maximum intensity. A student who confuses this with the destructive-interference condition (path difference a HALF-integer multiple of wavelength) picks A.

2. Q2. Unpolarised light of intensity I0 falls on two crossed ideal polaroids with a third ideal polaroid between them. The maximum intensity that can emerge, and the orientation of the middle polaroid that achieves it, are:

   • A.I0/8 with the middle axis at 45 degree to the first✓
   • B.I0/4 with the middle axis at 45 degree to the first
   • C.I0/2 with the middle axis parallel to the first
   • D.Zero for every orientation
   Solution

   With the middle axis at theta to the first, the output is (I0/2) cos^2 theta cos^2(90 degree - theta) = (I0/8) sin^2 2 theta. This is greatest when 2 theta = 90 degree, that is theta = 45 degree, giving I0/8.

3. Q3. In a double slit experiment the intensities from the two slits are 9I0 and I0. The fringe visibility, defined as (I_max − I_min)/(I_max + I_min), is:

   • A.0.8
   • B.0.6✓
   • C.1.0
   • D.0.4
   Solution

   Amplitudes are 3a and a. I_max ∝ (3a + a)^2 = 16a^2, I_min ∝ (3a − a)^2 = 4a^2. Visibility = (16 − 4)/(16 + 4) = 12/20 = 0.6. Equivalently 2√(I1 I2)/(I1 + I2) = 2×3/10 = 0.6.

4. Q4. For DESTRUCTIVE interference (a dark fringe) to occur at a point, the path difference must be:

   • A.An integer multiple of the wavelength specifically
   • B.A half-integer multiple of the wavelength ((n+1/2) x lambda, where n=0, 1, 2, and so on)✓
   • C.Always exactly zero
   • D.Exactly equal to the slit separation d
   Solution

   Destructive interference (producing a dark fringe/minimum) occurs when the path difference is a HALF-INTEGER multiple of the wavelength: path difference = (n+1/2) x lambda, where n=0, 1, 2, and so on -- at such points, the two waves arrive exactly OUT OF PHASE (180 degrees apart), and (for equal-amplitude waves) their amplitudes exactly cancel, producing minimum (ideally zero) intensity. A student who confuses this with the constructive-interference condition (integer multiples of the full wavelength) picks A.

5. Q5. The angular width of a fringe in Young's experiment performed in air is 0.20°. If the whole apparatus is immersed in water (refractive index 4/3), the angular fringe width becomes:

   • A.0.20°
   • B.0.27°
   • C.0.10°
   • D.0.15°✓
   Solution

   Angular fringe width θ = λ/d. In water the wavelength becomes λ/n while d is unchanged, so θ' = θ/n = 0.20° × 3/4 = 0.15°.

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